Geometric mean of fractional derivative

Question

What is the geometrical mean of the fractional derivative (of order $\alpha \in (0,1)$) for a function $f:\Bbb R \rightarrow \Bbb R$?
For example $f$ is increasing on $\Bbb R$ if $f'$ is positive.

Answer 1

Caputo fractional derivative for $\alpha \in (0,1)$ is

$\left(\frac{d}{dx}\right)^{\alpha}(f(t))$=$\frac{1}{\Gamma(1-\alpha)}$$\int_a^{t_0} $$\frac{f'(\tau)}{(t-\tau)^{\alpha}}$.

If $\frac{1}{\Gamma(1-\alpha)}\int_a^{t_0} \frac{f'(\tau)}{(t-\tau)^{\alpha}}>0$, then the plot of function $\frac{f'(\tau)}{(t-\tau)^{\alpha}}$ is on up side of $t$-axis for $\tau \in (a,t_0)$

On the other hand, if $f$ is increasing at $t=t_0$, then $f'(t)>0$ for $t \in(t_0-\epsilon,t_0+\epsilon) $. We assume that $a=t_0-\epsilon$ in definition of fractional derivative. Hence, since ${(t-\tau)^{\alpha}}$ is positive for $\tau \in (a,t_0)$, sign of $f'(t)$ is equal to sign of $\left(\frac{d}{dx}\right)^{\alpha}(f(t))$ at $t=t_0$